Question

use the probability distribution to complete parts (a) through (d) below.

the probability distribution of number of televisions per household in a small town
\\(\

$$\begin{array}{ccccc} x & 0 & 1 & 2 & 3 \\\\ p(x) & 0.02 & 0.16 & 0.28 & 0.54 \\end{array}$$

\\)

(a) find the probability of randomly selecting a household that has one or two televisions.
the probability is 0.44.
(type an integer or a decimal. do not round.)

(b) find the probability of randomly selecting a household that has two or more televisions.
the probability is 0.82.
(type an integer or a decimal. do not round.)

(c) find the probability of randomly selecting a household that has between one and three televisions, inclusive.
the probability is 0.98.
(type an integer or a decimal. do not round.)

(d) find the probability of randomly selecting a household that has at most two televisions.
the probability is
(type an integer or a decimal. do not round.)

Explanation:

Identify the relevant outcomes for "at most two"

$$ x \le 2 \implies x \in \{0, 1, 2\} $$

Sum the corresponding probabilities

$$ P(x \le 2) = P(0) + P(1) + P(2) $$

Calculate the final probability

$$ P(x \le 2) = 0.02 + 0.16 + 0.28 = 0.46 $$

Answer:

(d) Find the probability of randomly selecting a household that has at most two televisions.

The probability is <blank>0.46</blank>
(Type an integer or a decimal. Do not round.)